# How Do You Calculate Signal Propagation Velocity in Transmission Lines?

• lam58
In summary, the conversation discusses how to find the wave propagation velocity for an overhead line and underground cable using simplifying assumptions. The solution for the overhead line has been attempted, with values found for Z_0 and \gamma. However, the issue of finding the wave propagation velocity arises, with an attempt to use the assumption U = \frac{1}{\sqrt{LC}} and a resulting value of 725.5x10^3 ms^{-1}. The question of whether this assumption is allowed for a non-lossless line is also raised. The conversation also includes a discussion on the units used for Z and Y and their relation to ωL and ωC.
lam58
Hello, I'm stuck on how to find the propagation velocity of the signal in the Line as stated in Question C on the attached image below.

1. Homework Statement

Table one (in the attached image) shows typical values of Z and Y for an overhead line and underground cable. Please note that this is an engineering problem, thus all complex numbers are expressed in a+jb rather than a+ib.

## Homework Equations

Using simplifying assumptions calculate $$Z_0, \gamma$$ and the wave propagation velocity for each case.

## The Attempt at a Solution

So far I've only attempted the solution for the overhead line, I've found $$Z_0$$ and $$\gamma$$, however, I'm not entirely sure how to find the wave propagation velocity. I've attempted it, but to no avail.

$$Z_0 = \sqrt{\frac{Z}{Y}} = \sqrt{\frac{0.03 + j0.38}{0 + j5{x10^{-6}}}} = \sqrt{76{x10^3} \angle-4.5^o}$$

$$= 275.7\angle-2.25^o \Omega = 275.5 - j10.82 \Omega$$

$$\gamma = \sqrt{ZY} = \sqrt{(0.03 + j0.38)(0 + j5x10^{-6})} = \sqrt{(0.4 \angle85.5^o)(5x10^{-6} \angle90^o)}$$

$$= 1.41x10^{-3} \angle{87.75^o} = 5.5x10^{-5} + j1.41x10^{-3}$$

At this point I'm not sure how to get the wave propagation velocity. I've tried making the assumption that propagation velocity $$U = \frac{1}{\sqrt{LC}}$$ the only problem here is that this is for a lossless line and I'm not entirely sure if this assumption is allowed considering this isn't really a lossless line. In any case I get:

$$U = \frac{1}{\sqrt{\frac{(0.38)(5x10^{-6})}{1000}}} = 725.5x10^3 ms^{-1}$$ Note that I divided by 1000 because the stated values in the table are given per km.

Am I on the right track here or am I way off?

#### Attachments

• wave.png
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Aren't the imaginary parts of Z and Y equal to ωL and ωC, respectively? So, for example, 0.38 Ω/km is the value of ωL rather than the value of L. Or am I mistaken?

If you include the units given for Z and Y in the table when you calculate U, do you get m/s?

## 1. What is wave propagation velocity?

Wave propagation velocity refers to the speed at which a wave travels through a medium. It is a measure of how quickly the disturbance created by the wave moves through the medium.

## 2. How is wave propagation velocity calculated?

The wave propagation velocity can be calculated by dividing the wavelength of the wave by its period. It can also be calculated by multiplying the frequency of the wave by its wavelength.

## 3. What factors affect wave propagation velocity?

The velocity of a wave is influenced by the properties of the medium it travels through, such as density, elasticity, and temperature. The wavelength and frequency of the wave also play a role in determining its velocity.

## 4. What are some examples of wave propagation velocity in everyday life?

Examples of wave propagation velocity in everyday life include the speed of sound waves traveling through air, the speed of seismic waves traveling through the Earth's crust, and the speed of light waves traveling through a vacuum.

## 5. How is wave propagation velocity important in science?

Wave propagation velocity is an important concept in various fields of science, such as physics, engineering, and geology. It helps us understand how waves behave and how they can be used in various applications, such as communication, imaging, and energy transfer.

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